Consider triangles ADB and AEB.
1. The area of the triangle ADB is
[tex]A_{\triangle ADB}=\dfrac{1}{2}\cdot AB\cdot h_D.[/tex]
2. The area of the triangle AEB is
[tex]A_{\triangle AEB}=\dfrac{1}{2}\cdot AB\cdot h_E.[/tex]
3. Since AB and p are parallel lines, then
[tex]h_D=h_E.[/tex]
This gives you that
[tex]A_{\triangle ADB}=A_{\triangle AEB}=a.[/tex]
Therefore,
[tex]A_{\triangle ABC}=A_{\triangle ACE}+A_{\triangle AEB}=b+a.[/tex]
Answer: [tex]A_{\triangle ABC}=b+a.[/tex]